Problem 9 - Precalculus: Functions and Graphs

Question

Express as a polynomial. $$(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})$$

Step-by-Step Solution

Answer

The expression is \(x - y\).

Step 1: Identify the Expression Type

The given expression is in the form of \((a + b)(a - b)\), which is a difference of squares. Here, \(a = \sqrt{x}\) and \(b = \sqrt{y}\).

Step 2: Apply the Difference of Squares Formula

The difference of squares formula is \((a + b)(a - b) = a^2 - b^2\). Using \(a = \sqrt{x}\) and \(b = \sqrt{y}\), we apply it to get:\[(\sqrt{x})^2 - (\sqrt{y})^2\]

Step 3: Simplify the Squares

Square both \(\sqrt{x}\) and \(\sqrt{y}\): - \((\sqrt{x})^2 = x\) - \((\sqrt{y})^2 = y\)Therefore, the expression becomes:\[x - y\]

Step 4: Express as a Polynomial

The simplified expression \(x - y\) is already a polynomial. Hence, the expression \((\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})\) is expressed as the polynomial \(x - y\).

Key Concepts

Polynomial Expression

A polynomial expression is a mathematical expression consisting of variables and coefficients, combined using addition, subtraction, multiplication, and non-negative integer exponents. In general, polynomials can be expressed as sums of terms like \( ax^n \), where \( a \) is a coefficient and \( n \) is a whole number exponent. Polynomials can have one or more variables, but they don't include variables that are divided by another variable or have negative or fractional exponents.Our goal in the exercise is to express the given expression as a polynomial. A polynomial expression can simply be a single term like \( x \) or a combination such as \( x - y \). In the original problem, the expression

• \((\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})\)

represented a difference of squares, and through simplification, it became \( x - y \), which is a polynomial.

Simplification

Simplification is the process of transforming a mathematical expression into its simplest form. This often involves combining like terms, factoring expressions, or applying algebraic formulas. In our exercise, simplifying the expression helps us understand it as a polynomial.The given expression \((\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})\) needs simplification using the difference of squares formula:

• The difference of squares formula is \((a + b)(a - b) = a^2 - b^2\).
• Here, we identified \(a = \sqrt{x}\) and \(b = \sqrt{y}\).

By applying the formula, we simplify:

• \((\sqrt{x})^2 - (\sqrt{y})^2\)

which further reduces to

• \(x - y\).

This shows that the complex expression can be expressed in a simpler polynomial form.

Square Roots in Algebra

Square roots in algebra are often used to simplify expressions or solve equations involving squared variables. A square root \(\sqrt{x}\) is a number that, when multiplied by itself, gives the original number \(x\). Understanding square roots is vital when dealing with the simplification of expressions such as differences of squares.In the exercise,

• the expression \((\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})\) includes square root terms, \(\sqrt{x}\) and \(\sqrt{y}\).
• When squares of these terms are simplified using \((\sqrt{x})^2 = x\) and \((\sqrt{y})^2 = y\),

the expression transforms from having roots to simplified polynomial terms. Recognizing these transformations is key to understanding the behavior of algebraic expressions and using foundational operations such as the difference of squares effectively.